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I understand the concept of voltage and why electrons flow and my guess is their potential gradually decreases as they get closer to the cathode until it reaches 0. However, the energy in a circuit is "lost" (converted into heat/light) by resistance either within the wires or due to a device such as a lightbulb and I've been told the energy used up by these equals the energy put into the electrons at the beginning of the circuit (so they start off with an amount of energy and use up all of it by the end of their journey).

How does the circuit know how much energy to use? Let's say a 3V lightbulb is in a 3V circuit. We've learnt if we were to measure voltage around that lightbulb it'd be 3V. If we put in two 3V lightbulbs, each would get only 1.5V because they have to "share" the energy available. That makes no sense, because how would the electrons know when to use up how much energy?

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  • $\begingroup$ Part of your problem may be a misconception of what energy 'is'. I'm still learning myself. If you read Feynman's lectures you'll learn that really all we know about energy is it's this something that is always conserved [within a system]. But really we don't know more beyond that of what it is , fundamentally speaking. Nature tends to bring its states to an equilibrium; energy and entropy both measures of that equilibrium. How does the circuit (nature) know? That's a good question. $\endgroup$
    – docscience
    Sep 28, 2017 at 20:04
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    $\begingroup$ The circuit doesnt "know". Resistance essentially is similar to friction. Imagine you are pulling a train car (electrons) along rails (conductor). Your energy will be lost for friction (resistance). Now imagine you are pulling two connected cars (two lightbulbs in series) with the same force (voltage). You'd be moving slower and because of this would ultimately lose less energy (two lightbulbs in series use the total of a half the power of one, a quarter each). This is a very rough analogy, but it may give you a sense of how the energy is spent. $\endgroup$
    – safesphere
    Sep 28, 2017 at 20:17
  • $\begingroup$ @safesphere, that's a pretty good analogy. $\endgroup$ Sep 29, 2017 at 1:10
  • $\begingroup$ @safesphere Please don't use comments to leave answers, but write an actual answer instead. Comments are for requesting clarification or critiquing the question, not answering it. $\endgroup$
    – ACuriousMind
    Sep 29, 2017 at 1:43
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    $\begingroup$ Thanks to everyone who answered, or, well, left a comment :) it does all make more sense to me now, I just had to think about it a bit. The thing is in school we're told what happens but not how or why and my teacher doesn't want us to ask questions, which is a shame. But thanks again! $\endgroup$
    – llama
    Sep 29, 2017 at 18:21

3 Answers 3

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The state you describe is the steady state solution. It's what happens after a long period of time. Of course, in this case "long period of time" is measured in nanoseconds.

The electrical fields propagate at some reasonable fraction of the speed of light (2/3 c for copper wire). During this time, you do in fact find effects that look like "the electrons didn't know." Sometimes the fields will grow stronger in one area just because of inductive or capacitive effects with the environment. However, after a while, these effects will start to disappear as they system approaches its steady state. For a simple battery, resistor, and wire circuit, that may take 5 ns or less. (the speed of light is roughly 1 foot per nanosecond, if you're not afraid of English units).

This is similar to how things work in aerodynamics. The air "knows" about the airfoil that's upstream because there's pressure waves that flow upstream to "tell" the air about the incoming airfoil. These pressure waves build up fast enough that we can typically just ignore their development. We just assume they are there. However, once we get to supersonic speeds, the pressure waves can't reach upstream air before the airfoil gets there. At that point we get shock waves.

That scene can be turned around. If we imagine a wind tunnel with a stationary airfoil, the result is more similar to that of the electrical circuit. When you connect the circuit/activate the wind tunnel, a "shock wave" of current/air flows around the airfoil using non-steady-state flows. Shortly afterwards, the change in current/change in air starts to calm down, and the current/air "fills in" to whatever the steady state optimum is.

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Assuming a 100% resistive (ohmic) load, we have:

$$\text{V}_{\space\text{in}}=\text{V}_{\space\text{R}_1}+\text{V}_{\space\text{R}_2}\tag1$$

Now, using Ohm's law:

$$\text{V}_{\space\text{in}}=\text{I}_{\space\text{R}_1}\cdot\text{R}_1+\text{I}_{\space\text{R}_2}\cdot\text{R}_2\tag2$$

We know that:

  • $$\text{R}=\text{R}_1=\text{R}_2\tag3$$
  • $$\text{I}_{\space\text{in}}=\text{I}_{\space\text{R}_1}=\text{I}_{\space\text{R}_2}\tag4$$

So, we get:

$$\text{V}_{\space\text{in}}=\text{I}_{\space\text{in}}\cdot\text{R}+\text{I}_{\space\text{in}}\cdot\text{R}=2\cdot\text{I}_{\space\text{in}}\cdot\text{R}\space\Longleftrightarrow\space\text{I}_{\space\text{in}}=\frac{\text{V}_{\space\text{in}}}{2\cdot\text{R}}\tag5$$

And the energy used is:

$$\text{E}_{\space\text{in}}\left(t\right)=\int\text{P}_{\space\text{in}}\left(t\right)\space\text{d}t=\int\text{V}_{\space\text{in}}\left(t\right)\cdot\text{I}_{\space\text{in}}\left(t\right)\space\text{d}t\tag6$$

So, your problem:

$$\text{E}_{\space\text{in}}\left(t\right)=\int\text{V}_{\space\text{in}}\cdot\text{I}_{\space\text{in}}\space\text{d}t=$$ $$\int\text{V}_{\space\text{in}}\cdot\frac{\text{V}_{\space\text{in}}}{2\cdot\text{R}}\space\text{d}t=\frac{\text{V}_{\space\text{in}}^2}{2\cdot\text{R}}\int1\space\text{d}t=\frac{\text{V}_{\space\text{in}}^2}{2\cdot\text{R}}\cdot t+\text{K}\tag7$$

Assuming that $\text{K}=0$, then we get:

$$\text{E}_{\space\text{in}}\left(t\right)=\frac{3^2}{2\cdot\text{R}}\cdot t=\frac{9}{2}\cdot\frac{t}{\text{R}}\tag8$$

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Think of voltage as a pressure difference.

If you squeeze a water hose so that only little water comes out, then that water has dropped from the high pressure before the squeeze to a lower pressure after. The entire pressure difference is "spent".

If you squeeze at two different points, then the same happens around the first squeeze initially, but after that the water can come no further. Because there is one more squeeze. Water will thus build up between the squeezes, building up a new pressure here. As this intermediate pressure is increases, more and more water is pressed through the second squeeze. Soon equally much water is pressed through the second squeeze as is entering the intermediate zone.

The pressure difference between either side of the first squeeze decreases during the accumulation on this intermediate zone. And the pressure difference between either side of the second squeeze increases. But nevertheless there is a total pressure difference across both squeezes equal to the water pressure from the tab - it can never be any higher or lower than that in total.

And here you have it. The total available pressure difference (voltage) was "shared" between the resistances (in series) due to water (charge) accumulating on the right places until a steady flow has been reached again.

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